Great! Let's break down the problem and find the remainder when the polynomial [tex]\( P(x) \)[/tex] is divided by the quadratic polynomial [tex]\((x-a)(x-b)\)[/tex].
### Step-by-Step Solution:
#### Step 1: Understand the Remainder Theorem for Quadratic Divisors
Given [tex]\( P(x) \)[/tex] is divided by [tex]\( (x-a)(x-b) \)[/tex], the remainder [tex]\( R(x) \)[/tex] will be a linear polynomial of the form [tex]\( R(x) = C_1 x + C_0 \)[/tex], where [tex]\( C_1 \)[/tex] and [tex]\( C_0 \)[/tex] are constants to be determined.
#### Step 2: Express [tex]\( P(x) \)[/tex] in Terms of the Divisor and the Remainder
We can write the division as:
[tex]\[ P(x) = (x-a)(x-b) Q(x) + R(x) \][/tex]
Here, [tex]\( Q(x) \)[/tex] is the quotient, and [tex]\( R(x) = C_1 x + C_0 \)[/tex] is the remainder.
#### Step 3: Use the Fact that Remainder Holds at Specific Points
Given the definition of the remainder, we can substitute [tex]\( x = a \)[/tex] and [tex]\( x = b \)[/tex] to find the unknowns [tex]\( C_1 \)[/tex] and [tex]\( C_0 \)[/tex]:
1. At [tex]\( x = a \)[/tex]:
[tex]\[ P(a) = (a-a)(a-b)Q(a) + R(a) \][/tex]
[tex]\[ P(a) = 0 + R(a) \][/tex]
So,
[tex]\[ R(a) = P(a) \][/tex]
Since [tex]\( R(x) = C_1 x + C_0 \)[/tex], substituting [tex]\( x = a \)[/tex] gives:
[tex]\[ C_1 a + C_0 = P(a) \quad \text{(1)} \][/tex]
2. At [tex]\( x = b \)[/tex]:
[tex]\[ P(b) = (b-a)(b-b)Q(b) + R(b) \][/tex]
[tex]\[ P(b) = 0 + R(b) \][/tex]
So,
[tex]\[ R(b) = P(b) \][/tex]
Similarly, substituting [tex]\( x = b \)[/tex] into [tex]\( R(x) = C_1 x + C_0 \)[/tex]:
[tex]\[ C_1 b + C_0 = P(b) \quad \text{(2)} \][/tex]
#### Step 4: Solve for [tex]\( C_0 \)[/tex] and [tex]\( C_1 \)[/tex]
From equations (1) and (2), we have:
[tex]\[ C_1 a + C_0 = P(a) \][/tex]
[tex]\[ C_1 b + C_0 = P(b) \][/tex]
Subtract the first equation from the second:
[tex]\[ (C_1 b + C_0) - (C_1 a + C_0) = P(b) - P(a) \][/tex]
[tex]\[ C_1 (b - a) = P(b) - P(a) \][/tex]
[tex]\[ C_1 = \frac{P(b) - P(a)}{b - a} \][/tex]
Now, solve for [tex]\( C_0 \)[/tex] using equation (1):
[tex]\[ C_1 a + C_0 = P(a) \][/tex]
[tex]\[ \left(\frac{P(b) - P(a)}{b - a}\right) a + C_0 = P(a) \][/tex]
[tex]\[ C_0 = P(a) - \left(\frac{P(b) - P(a)}{b - a}\right) a \][/tex]
#### Step 5: Form the Remainder Expression
Now we have the values of [tex]\( C_1 \)[/tex] and [tex]\( C_0 \)[/tex]:
[tex]\[ C_1 = \frac{P(b) - P(a)}{b - a} \][/tex]
[tex]\[ C_0 = P(a) - \left(\frac{P(b) - P(a)}{b - a}\right) a \][/tex]
So, the remainder [tex]\( R(x) \)[/tex] is:
[tex]\[ R(x) = C_1 x + C_0 \][/tex]
[tex]\[ R(x) = \left( \frac{P(b) - P(a)}{b - a} \right) x + \left( P(a) - \left( \frac{P(b) - P(a)}{b - a} \right) a \right) \][/tex]
[tex]\[ R(x) = \left( \frac{P(b) - P(a)}{b - a} \right) x - \left( \frac{P(b) - P(a)}{b - a} \right) a + P(a) \][/tex]
[tex]\[ R(x) = \left( \frac{P(b) - P(a)}{b - a} \right) (x - a) + P(a) \][/tex]
Therefore, the remainder when [tex]\( P(x) \)[/tex] is divided by [tex]\( (x-a)(x-b) \)[/tex] is:
[tex]\[ \left( \frac{P(b) - P(a)}{b - a} \right) (x - a) + P(a) \][/tex]
This completes the proof.
